First slide

Introduction to limits

Question

$lim_{x \to 0} \frac{\sqrt{1 - \cos x}}{x}$ is equal to

Moderate

A

$- \frac{1}{\sqrt{2}}$
B

$\frac{1}{\sqrt{2}}$
C

0
D

does not exist

Solution

$\sqrt{1 - \cos x} = \{\begin{cases} - \sqrt{2} \sin \frac{x}{2}, x < 0 \\ \sqrt{2} \sin \frac{x}{2}, x \geq 0 \end{cases}$
Therefore,
$\begin{array}{l} lim_{x \to 0^{-}} \frac{\sqrt{1 - \cos x}}{x} = lim_{x \to 0^{-}} \frac{- \sqrt{2} \sin \frac{x}{2}}{x} = - \frac{1}{\sqrt{2}} . \\ lim_{x \to 0^{+}} \frac{\sqrt{1 - \cos x}}{x} = lim_{x \to 0^{+}} \frac{\sqrt{2} \sin \frac{x}{2}}{x} = \frac{1}{\sqrt{2}} \end{array}$
since,
$\begin{array}{l} lim_{x \to 0^{-}} \frac{\sqrt{1 - \cos x}}{x} \neq lim_{x \to 0^{+}} \frac{\sqrt{1 - \cos x}}{x} \\ lim_{x \to 0} \frac{\sqrt{1 - \cos x}}{x} does not exist. \end{array}$

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